CBSE Class 8 Mathematics The Baudhayana Pythagoras Theorem Short Answer Questions with Answers

Short Answer Questions

Question 1. A square sheet has sidelength 6 cm. A second square is made by doubling the sidelength.
(i) Find the area of each square.
(ii) How many times the first area is the second area? Explain why doubling the side does not double the area.
Answer:
(i) First square: 6 × 6 = 36 cm². Second square: side 12 cm, area 12 × 12 = 144 cm².
(ii) 144 ÷ 36 = 4, so the area becomes 4 times, not 2 times.
Area = side × side, so when the side is multiplied by 2, the area is multiplied by 2 × 2 = 4.
To double the area, the new square must be built on the diagonal, not on a doubled side.
In simple words: Doubling the side multiplies the area by 4 (36 cm² becomes 144 cm²), not by 2.

Teacher's Note:
1. Both the length and the breadth get doubled.
2. Compare the areas by dividing.

Question 2. State Baudhāyana’s rule from Śulba-Sūtra Verse 1.9 about the diagonal of a square. Using it, write the area of the square drawn on the diagonal of a square of sidelength 5 units, and say how many of the small congruent triangles make up each of the two squares.
Answer:
Verse 1.9: the diagonal of a square produces a square of double the area of the original square.
Original square: area 5 × 5 = 25 sq. units.
Square on its diagonal: area 2 × 25 = 50 sq. units.
The east-west and north-south lines cut the original square into 2 congruent small triangles, while the square on the diagonal is made of 4 such triangles. 4 is twice 2, so the area is doubled.
In simple words: The square on the diagonal of a 5-unit square has area 50, made of 4 small triangles against the original 2.

Teacher's Note:
1. Count triangles to compare areas.
2. The diagonal of this square is √50 units.

Question 3. A square paper of sidelength 8 cm is folded inwards so that the crease lines pass through the midpoints of its sides, giving an inner square PQRS.
(i) Find the area of PQRS.
(ii) Find its sidelength in exact form.
Answer:
(i) The four corner triangles cut off by the creases together equal the inner figure, so PQRS is half the paper.
Area of PQRS = \( \frac{1}{2} \) × 8 × 8 = 32 cm².
(ii) If s is the side of PQRS, then s² = 32, so s = √32 = 4√2 cm (between 5.6 cm and 5.7 cm).
In simple words: PQRS has half the area of the 64 cm² paper, so 32 cm², and its side is √32 = 4√2 cm.

Teacher's Note:
1. Midpoint square = half area.
2. Exact form keeps the square root.

Question 4. The two equal sides of an isosceles right triangle are each of length 7 units. Find the exact length of the hypotenuse, and give bounds for it with one digit after the decimal point.
Answer:
For an isosceles right triangle, c² = 2a².
Here a = 7, so c² = 2 × 49 = 98.
Therefore c = √98 = 7√2 units.
Bounds: 9.8² = 96.04 and 9.9² = 98.01, and 96.04 < 98 < 98.01.
So 9.8 < √98 < 9.9.
In simple words: The hypotenuse is √98 = 7√2 units, which lies between 9.8 and 9.9.

Teacher's Note:
1. Start from 9² = 81 and 10² = 100, then refine.
2. Keep the exact value and the bounds separate.

Question 5. Explain, using areas of squares, why √2 must lie between 1 and 2. Then state the closer bounds 1.4 and 1.5 and justify them with one calculation each.
Answer:
A square of side √2 has area 2 sq. units.
A square of side 1 has area 1, and 1 < 2, so 1 < √2.
A square of side 2 has area 4, and 2 < 4, so √2 < 2. Hence 1 < √2 < 2.
Closer bounds: 1.4² = 1.96, which is less than 2; 1.5² = 2.25, which is more than 2. So 1.4 < √2 < 1.5.
In simple words: Squares of side 1 and 2 have areas 1 and 4, which trap 2; and 1.4² = 1.96, 1.5² = 2.25 trap 2 more closely.

Teacher's Note:
1. A bigger square always has a bigger side.
2. Each check needs only one multiplication.

Question 6. State the Baudhāyana–Pythagoras theorem for a right-angled triangle with sides a, b and hypotenuse c. Use it to find the hypotenuse of a right-angled triangle whose shorter sides are 9 units and 12 units.
Answer:
Theorem: If a right-angled triangle has sides a, b and hypotenuse c, then a² + b² = c².
Here a = 9 and b = 12, so c² = 81 + 144 = 225.
Therefore c = √225 = 15 units.
In simple words: a² + b² = c², so the hypotenuse is √(81 + 144) = 15 units.

Teacher's Note:
1. (9, 12, 15) = 3 × (3, 4, 5).
2. Square, add, then take the square root.

Question 7. What is a Baudhāyana triple? Verify that (8, 15, 17) is one, and state with a reason whether it is primitive.
Answer:
A Baudhāyana triple is a triple (a, b, c) of positive integers that forms the sidelengths of a right triangle, that is, a² + b² = c².
Check: 8² + 15² = 64 + 225 = 289 and 17² = 289. The two are equal, so (8, 15, 17) is a Baudhāyana triple.
It is primitive, because 8, 15 and 17 have no common factor greater than 1.
In simple words: (8, 15, 17) works since 64 + 225 = 289 = 17², and it is primitive as the numbers share no common factor.

Teacher's Note:
1. 17 is prime and does not divide 8 or 15.
2. This triple is in Baudhāyana’s own list.

Question 8. Fermat studied the equation xⁿ + yⁿ = zⁿ for natural numbers x, y, z.
(i) What did Fermat claim for n > 2?
(ii) Who finally proved this claim, and in which year?
Answer:
(i) Fermat claimed that for n > 2 the equation has no solution in natural numbers: no cube is the sum of two cubes, no fourth power is the sum of two fourth powers, and so on.
(ii) It was finally proved by Andrew Wiles in 1994, more than 300 years after Fermat wrote it in the margin of his book.
In simple words: Fermat said xⁿ + yⁿ = zⁿ has no natural-number solution for n > 2, and Andrew Wiles proved it in 1994.

Teacher's Note:
1. For n = 2 the solutions are Baudhāyana triples.
2. The claim stayed unproved for over 300 years.

Question 9. The hypotenuse of an isosceles right triangle is √98 units. Using the relation between the hypotenuse and the equal sides, find the length of each of the other two sides. Show the steps.
Answer:
For equal sides a and hypotenuse c, c² = 2a².
Given c = √98, so c² = 98.
Then 98 = 2a², giving a² = \( \frac{98}{2} \) = 49.
So a = √49 = 7. Each of the other two sides is 7 units long.
In simple words: 98 = 2a² gives a² = 49, so each equal side is 7 units.

Teacher's Note:
1. Square the hypotenuse to remove the root.
2. Check: 7² + 7² = 98.

Question 10. A right-angled triangle has one shorter side of length 20 cm and a hypotenuse of length 29 cm. Find the third side, and state the Baudhāyana triple formed by the three sides.
Answer:
Let the unknown side be b. Then 20² + b² = 29².
400 + b² = 841, so b² = 841 − 400 = 441.
b = √441 = 21 cm.
The sides give the Baudhāyana triple (20, 21, 29).
In simple words: b² = 841 − 400 = 441, so the third side is 21 cm and the triple is (20, 21, 29).

Teacher's Note:
1. Subtract to find a shorter side.
2. (20, 21, 29) is primitive.

Question 11. Starting from the primitive triple (5, 12, 13), write the scaled triple obtained with k = 4. Verify by calculation that it satisfies a² + b² = c², and say whether the scaled triple is primitive.
Answer:
Scaling (5, 12, 13) by k = 4 gives (20, 48, 52).
20² + 48² = 400 + 2304 = 2704, and 52² = 2704.
Since the two are equal, (20, 48, 52) is a Baudhāyana triple.
It is not primitive, because all three numbers have the common factor 4.
In simple words: 4 × (5, 12, 13) = (20, 48, 52) satisfies 400 + 2304 = 2704 = 52², but it is not primitive.

Teacher's Note:
1. Scaling multiplies each square by k².
2. Any scaled triple with k > 1 has the common factor k.

Question 12. The number 49 is an odd square.
(i) Find n such that 49 is the nth odd number.
(ii) Use (n − 1)² + (2n − 1) = n² to write the Baudhāyana triple it produces, and check it.
Answer:
(i) The nth odd number is 2n − 1. So 2n − 1 = 49 gives 2n = 50, that is n = 25.
(ii) Putting n = 25: 24² + 49 = 25².
49 = 7², so 24² + 7² = 25².
Check: 576 + 49 = 625 = 25². The triple is (7, 24, 25).
In simple words: 49 is the 25th odd number, which gives 24² + 7² = 25², the triple (7, 24, 25).

Teacher's Note:
1. The odd square itself becomes the square of a side.
2. The other side and the hypotenuse differ by 1.

Question 13. A rectangular photo frame measures 9 cm by 40 cm. Find the length of its diagonal, and explain in one line which right-angled triangle you used.
Answer:
The two sides of a rectangle meet at a right angle, so the diagonal is the hypotenuse of a right-angled triangle with shorter sides 9 cm and 40 cm.
d² = 9² + 40² = 81 + 1600 = 1681.
d = √1681 = 41 cm. The triple used is (9, 40, 41).
In simple words: The diagonal is √(81 + 1600) = 41 cm, using the right triangle formed by two sides and the diagonal.

Teacher's Note:
1. 41 × 41 = 1681.
2. (9, 40, 41) is a primitive triple.

Question 14. The diagonals of a rhombus are 16 units and 30 units long. Find the sidelength of the rhombus, stating the property of the diagonals that you use.
Answer:
The diagonals of a rhombus bisect each other at right angles.
Half-diagonals: \( \frac{16}{2} \) = 8 units and \( \frac{30}{2} \) = 15 units.
These are the shorter sides of a right triangle whose hypotenuse is a side of the rhombus.
Side = √(8² + 15²) = √(64 + 225) = √289 = 17 units.
In simple words: Half-diagonals 8 and 15 form a right triangle, so the side is √289 = 17 units.

Teacher's Note:
1. Halve both diagonals first.
2. (8, 15, 17) appears again.

Question 15. Describe, in steps, how a square of area exactly three times a given square can be constructed using only Baudhāyana’s two constructions (the square on a diagonal, and the combining of two different squares). Take the given square to have area 1 square unit.
Answer:
Start with the given square of area 1 sq. unit.
Step 1 (doubling): Build the square on its diagonal. By Verse 1.9 this square has area 2 sq. units, and its side is √2.
Step 2 (combining): Make a right-angled triangle whose perpendicular sides are √2 (side of the area-2 square) and 1 (side of the given square).
The square on its hypotenuse has area 2 + 1 = 3 sq. units, so the required square is built on that hypotenuse.
In simple words: Double the square using its diagonal, then combine that area-2 square with the original to get area 3.

Teacher's Note:
1. Areas add when squares are combined.
2. The side of the final square is √3.

Question 16. A classmate says, “√2 = 1.4142135, because that is what my calculator shows.” Explain in 2–3 points why no terminating decimal can have square exactly 2.
Answer:
The calculator value is only an approximation. Reasons:
1. A terminating decimal must end in some non-zero last digit.
2. When such a decimal is squared, the square also ends in a non-zero digit after the decimal point, so it can never be exactly 2 or 2.000…
3. Hence the decimal form of √2 never stops: √2 = 1.41421356…, and 1.4142135² is close to 2 but not equal to 2.
In simple words: A decimal that stops always has a square with a non-zero last decimal digit, so it can never square to exactly 2.

Teacher's Note:
1. The last digit of a square comes from the last digit of the number.
2. Calculators show only a few digits of an endless decimal.

Question 17. Show that √2 cannot be written as a fraction \( \frac{m}{n} \), where m and n are counting numbers. Use the fact that in the prime factorisation of a square number every prime occurs an even number of times.
Answer:
Suppose, if possible, that √2 = \( \frac{m}{n} \) with m, n counting numbers.
Squaring both sides: 2 = \( \frac{m^2}{n^2} \), so 2n² = m².
In n² and in m², the prime 2 occurs an even number of times.
On the left side, 2n² has one extra factor 2, so the prime 2 occurs an odd number of times there, while on the right it occurs an even number of times.
The same number cannot have both, so √2 cannot be written as \( \frac{m}{n} \). (This proof is due to Euclid, c. 300 BCE.)
In simple words: If √2 = m/n then 2n² = m², but the prime 2 would appear an odd number of times on one side and an even number on the other.

Teacher's Note:
1. This is a proof by contradiction.
2. Count only the factor 2 on each side.

Question 18. A lotus stands in a pond with the tip of its stem 2 units above the water. A breeze pushes the tip down until it just touches the water at a point 8 units away from the original position. Taking the depth of the water as x, form an equation and find the depth.
Answer:
Depth = x units. The stem sticks out 2 units, so the whole stem is x + 2 units.
The upright stem, the water surface and the pushed-over stem form a right triangle with shorter sides x and 8, and hypotenuse x + 2.
8² + x² = (x + 2)²
64 + x² = x² + 4x + 4
64 = 4x + 4, so 60 = 4x and x = 15.
Depth of the water = 15 units. Check: 8² + 15² = 289 = 17², and 17 = 15 + 2.
In simple words: Using 8² + x² = (x + 2)², the depth of the water is 15 units.

Teacher's Note:
1. This is the Līlāvatī lotus problem with new numbers.
2. The x² terms cancel, leaving a simple equation.

Question 19. An equilateral triangle has sidelength 10 units. Using the fact that an altitude bisects the opposite side, find the exact height and the exact area of the triangle.
Answer:
The altitude splits the triangle into two congruent right triangles, so it bisects the base.
Half of the base = \( \frac{10}{2} \) = 5 units.
The altitude h is a shorter side of a right triangle with hypotenuse 10 and other side 5: h² = 10² − 5² = 100 − 25 = 75.
So h = √75 = 5√3 units.
Area = \( \frac{1}{2} \) × 10 × 5√3 = 25√3 sq. units (about 43.3 sq. units).
In simple words: The height is √(100 − 25) = 5√3 units and the area is ½ × 10 × 5√3 = 25√3 sq. units.

Teacher's Note:
1. The side of the triangle is the hypotenuse.
2. √75 = √(25 × 3) = 5√3.

Question 20. A triple (a, b, c) is non-primitive and the three numbers have a common factor f > 1.
(i) Justify with algebra that \( \left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right) \) is also a Baudhāyana triple.
(ii) Check your reasoning on (15, 36, 39) with f = 3.
Answer:
(i) Write a = fp, b = fq, c = fr, where p = \( \frac{a}{f} \), q = \( \frac{b}{f} \), r = \( \frac{c}{f} \) are positive integers.
a² + b² = c² gives f²p² + f²q² = f²r², that is, f²(p² + q²) = f²r².
Dividing both sides by f² (not zero): p² + q² = r². So \( \left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right) \) is a Baudhāyana triple.
(ii) (15, 36, 39) with f = 3 gives (5, 12, 13). Check: 5² + 12² = 25 + 144 = 169 = 13². It is a Baudhāyana triple, and in fact a primitive one.
In simple words: Dividing a triple by a common factor f divides each square by f², so the relation a² + b² = c² still holds.

Teacher's Note:
1. Taking out f² from both sides is the key step.
2. This is the reverse of scaling.